处理连续行计算

Question

假设以下情况：

• 第一周：
  • 以前的案件
  • 10个新案件
  • 3宗已解决案件

​

• 第二周：
  • 7起以前的案件
  • 13个新案件
  • 15个已解决案件

​

• 第三周：
  • 先前5起案件
  • 6个新案件
  • 7个已结案的案件

​

此信息存储在以下类型的恢复表中：

RESUME_TABLE:
WEEK    | TOTAL_NEW |   TOTAL_SOLVED
1       |   10      |   3
2       |   13      |   15
3       |   6       |   7

我很难构建一个查询来获得以下结果：

REPORT_TABLE:
WEEK    |   PREV_TOTAL  |   NEW_CASES   |   SOLVED_CASES    |   NEW_TOTAL
1       |   0           |   10          |   3               |   7
2       |   7           |   13          |   15              |   5
3       |   5           |   6           |   7               |   4

这个想法看起来很琐碎，NEW_TOTAL = PREV_TOTAL + NEW_CASES - SOLVED_CASES，尽管我一直在为将PREV_TOTAL带到下一行的想法而苦苦挣扎。

我试图使用RESUME表(Oracle11g)上的一个视图来完成这个任务。

有人能帮我做一些示例代码吗？

Answer 1

非常简单和简洁的分析功能：

12:57:06 HR@vm_xe> l                                                                    
  1  select week                                                                        
  2         ,lag(total_cases_by_now - total_solved_by_now) over (order by week) prev_total
  3         ,total_new new_cases                                                        
  4         ,total_solved solved_cases                                                  
  5         ,total_cases_by_now - total_solved_by_now new_total                         
  6    from (                                                                           
  7    select week                                                                      
  8           ,total_new                                                                
  9           ,total_solved                                                             
 10           ,sum(total_new) over(order by week asc) as total_cases_by_now             
 11           ,sum(total_solved) over (order by week asc) as total_solved_by_now        
 12      from resume_table                                                              
 13* )                                                                                  
12:57:07 HR@vm_xe> /                                                                    

      WEEK   PREV_TOTAL  NEW_CASES SOLVED_CASES  NEW_TOTAL                                
---------- ------------ ---------- ------------ ----------                                
         1                      10            3          7                                
         2            7         13           15          5                                
         3            5          6            7          4                                

3 rows selected.                                                                        

Elapsed: 00:00:00.01                                                                    

Answer 2

您可以使用模型子句来解决这个问题：

with resume_table as
(
    select 1 week, 10 total_new, 3 total_solved from dual union all
    select 2 week, 13 total_new, 15 total_solved from dual union all
    select 3 week, 6 total_new,  7 total_solved from dual
)
select week, prev_total, total_new, total_solved, new_total
from resume_table
model
    dimension by (week)
    measures (total_new, total_solved, 0 prev_total, 0 new_total)
    rules sequential order
    (
        new_total[any] order by week = 
            nvl(new_total[cv(week)-1], 0) + total_new[cv()] - total_solved[cv()]
        ,prev_total[any] order by week = nvl(new_total[cv(week)-1], 0)
    )
order by week;

尽管这使得人们认为周总是一个连续的数字。如果这不是真的，您将需要添加一个row_number()。否则，-1可能不会引用前一个值。

看这个SQL Fiddle。

Answer 3

在RESUME_TABLE中添加一列(或创建一个视图，我认为这可能更好)：

RESUME_LEFT
WEEK | LEFT
1    | 7
2    | -2
3    | -1

就像这样：

CREATE VIEW resume_left
  (SELECT week,total_new-total_solved "left" FROM resume_table)

所以在REPORT_TABLE中，您可以有这样的定义：

PREV_TOTAL=(SELECT sum(left) FROM RESUME_LEFT WHERE week<REPORT_TABLE.week)

编辑

好的，这个视图是不必要的：

PREV_TOTAL=(SELECT sum(total_new)-sum(total_solved)
  FROM resume_table
  WHERE week<REPORT_TABLE.week)